Recent content by kevinsweeney

yes, it must be true for EVERY 2 pairs to which it applies to

...No, its not {x,y,z} in the set(as that would imply we could take any 3 numbers of our choice!), its the elements linked by "y" with (x,y) (y,z)...y just has to be in the 2nd position in the first set, and 1st position in the second set. There is actually another example in that set...

A) But it does apply for the whole set. There is no other terms fitting the criteria of (x,y)(y,z) only (1,2)(2,1) as you can see y is 2 in both. Can you show me another (x,y)(y,z) in the set? k) we came across that logic in truth tables if you remember, we had let's say x implies y...

A) r2 is transitive as for (x,y)[1,2](y,z)[2,1] there is a (x,z)[1,1] k) r5 is transiitve as there is no terms For (x,y)(yz) implying (x,z), therefore the theory is true because we can't prove it false(maths logic) everything else looks right

i can't see your answers, so can't tell where you are going wrong? Can you put up a screeshot/grab oh sorry, they are your answers, right?

D is not reflexive as 0(0) for (x,x) is not positive. E is reflexive as x -x will always be 0. 0 is a multiple of EVERY number. yes, b is reflexive as (0)0 is always >= 0.And how about q3?

I got these sorted. What part of q2 are you stuck with?

Determine which of these relations are reflexive. The variables x, y, x', y' represent integers. A. x∼y if and only if xy is positive. B. x∼y if and only if x+y is even. C. x∼y if and only if x−y is a multiple of 10. D. x∼y if and only if x−y is positive. E. x∼y if and only if x+y is...

Suppose that R1={(2,2),(2,3),(2,4),(3,2),(3,3),(3,4)}, R2={(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)}, R3={(2,4),(4,2)} , R4={(1,2),(2,3),(3,4)}, R5={(1,1),(2,2),(3,3),(4,4)}, R6={(1,3),(1,4),(2,3),(2,4),(3,1),(3,4)}, Determine which of these statements are correct. Check ALL correct answers...